AP Chemistry · Lesson 3

Reading a titration curve

Know Know it

In a titration, a solution of known concentration (the titrant) is added until it has reacted exactly with the sample. At the equivalence point, the moles of titrant match the moles of analyte by the coefficients; for a monoprotic acid and NaOH, MaVa=MbVbM_a V_a = M_b V_b.

Strong acid with strong base: the pH starts low, rises slowly, jumps sharply near the equivalence point and is 7.00 there at 25 °C.

Weak acid with strong base: the pH starts higher than for a strong acid of the same concentration. Then comes a gently rising buffer region, where HA and A−^- are both present. At the half-equivalence point, [HA]=[A−][\text{HA}] = [\text{A}^-], so pH=pKa\text{pH} = \text{p}K_a. At the equivalence point the solution holds A−^-, a weak base, so the pH is above 7. After it, excess hydroxide sets the pH.

Weak base with strong acid is the mirror image: the equivalence point is below pH 7. A polyprotic acid shows one equivalence point per acidic hydrogen when the KaK_a values are far apart; the pH halfway to the first is close to pKa1\text{p}K_{a1}, and the pH halfway between the first and second is close to pKa2\text{p}K_{a2}.

An indicator should change color in a pH range that includes the equivalence-point pH.

Worked example Easy

A 25.0 mL sample of 0.120 M monoprotic acid HA is titrated with 0.150 M NaOH. What volume of NaOH is needed to reach the equivalence point?

  1. 31.3 mL
  2. 25.0 mL
  3. 10.0 mL
  4. 20.0 mLAnswer

How to solve it

  1. Moles of acid: 0.0250×0.120=3.00×10−30.0250 \times 0.120 = 3.00 \times 10^{-3} mol.
  2. At equivalence, moles of NaOH = moles of HA (1:1).
  3. Volume of NaOH: 3.00×10−3/0.150=0.02003.00 \times 10^{-3} / 0.150 = 0.0200 L, or 20.0 mL.

Why each choice is right or wrong

  • A. Incorrect. 31.3 mL inverts the concentrations, 25.0×0.150/0.12025.0 \times 0.150/0.120; the more concentrated base needs less volume, not more.
  • B. Incorrect. 25.0 mL assumes equal volumes react, which is true only when the two concentrations are equal.
  • C. Incorrect. 10.0 mL is the half-equivalence volume, where only half of the acid has reacted.
  • D. Correct. Moles of HA = (0.0250 L)(0.120 M)=3.00×10−3(0.0250 \text{ L})(0.120 \text{ M}) = 3.00 \times 10^{-3} mol, and 3.00×10−3/0.150=0.02003.00 \times 10^{-3}/0.150 = 0.0200 L = 20.0 mL.

Trap Equivalence is not neutral

For a weak acid titrated with a strong base, the equivalence point is basic, because the solution then holds the conjugate base. Do not set pH 7 at the equivalence point unless both the acid and the base are strong. And the half-equivalence point, not the equivalence point, is where pH equals pKa\text{p}K_a.

Explain Explain it

Task verbs: determine, explain

To find the pH anywhere on a weak-acid curve, first do the stoichiometry in moles. Before equivalence, the mixture is a buffer: use the Henderson-Hasselbalch equation. At equivalence, use Kb=Kw/KaK_b = K_w/K_a for A−^- in the total volume. After equivalence, the excess hydroxide divided by the total volume sets the pOH.

Reading a curve backward is just as useful: the equivalence volume gives the moles of acid, and the pH at half that volume gives the pKa\text{p}K_a.

Worked example Hard

A 25.00 mL sample of 0.100 M acetic acid (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}) is titrated with 0.100 M NaOH. What is the pH after 30.00 mL of NaOH has been added?

  1. 12.22
  2. 8.72
  3. 11.96Answer
  4. 2.04

How to solve it

  1. Moles of acid: 0.02500×0.100=2.50×10−30.02500 \times 0.100 = 2.50 \times 10^{-3} mol. Moles of NaOH: 0.03000×0.100=3.00×10−30.03000 \times 0.100 = 3.00 \times 10^{-3} mol.
  2. Past equivalence, the excess is 5.0×10−45.0 \times 10^{-4} mol of OH−\text{OH}^- in 55.00 mL: [OH−]=9.1×10−3[\text{OH}^{-}] = 9.1 \times 10^{-3} M.
  3. pOH = 2.04, so pH = 14.00 - 2.04 = 11.96. The acetate made at equivalence adds very little hydroxide next to this excess.

Why each choice is right or wrong

  • A. Incorrect. A pH of 12.22 divides the excess hydroxide by 30.00 mL only; the total volume is 55.00 mL.
  • B. Incorrect. A pH of 8.72 is the pH at the equivalence point, 25.00 mL, but here 5.00 mL of excess NaOH has been added.
  • C. Correct. Excess hydroxide is 3.00×10−3−2.50×10−3=5.0×10−43.00 \times 10^{-3} - 2.50 \times 10^{-3} = 5.0 \times 10^{-4} mol in 55.00 mL, so [OH−]=9.1×10−3[\text{OH}^{-}] = 9.1 \times 10^{-3} M, pOH 2.04 and pH 11.96.
  • D. Incorrect. A pH of 2.04 is the pOH of the mixture, not its pH.

Explain Your turn

A student titrates a weak acid HA with NaOH and finds that the pH at the equivalence point is 8.7. Explain why the pH at the equivalence point is above 7. Include a chemical equation.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Names A−^-, the conjugate base of the weak acid, as a species present at the equivalence point
  • Writes the reaction of A−^- with water that makes hydroxide

Model answer

At the equivalence point all of the HA has been converted to A−^-, so the solution contains A−^-, Na+\text{Na}^+ and water. A−^- is the conjugate base of a weak acid, so it reacts with water: A−+H2O⇌HA+OH−\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-. The hydroxide it makes is more than the hydronium, so the pH is above 7.

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